Calculating Max. Demand Kilovolt-Amperes and Required Cable Current Rating
Calculating Max. Demand Kilovolt-Amperes and Required Cable Current Rating

A building site and electrical equipment

A building site is to have the following electrical equipment available for use:

  • Tower crane, electric motors totalling 75 kW at 415 V;
  • Sump pump, 5 kW at 240 V;
  • 60 tungsten lamps of 100 W each at 240 V;
  • 12 flood lamps of 400 W each at 240 V;
  • 20 hand tools of 400 W at 110 V.

To calculate //

  1. Find the total kilovolt-amperes to be supplied to the site if the power factor of all rotary equipment is 0.8.
  2. Find the electrical current rating for the incoming supply cable to the site.
  3. Estimate the cost of electricity consumed on the site during a 12-month contract.

Ok, let’s do it…

For rotating machinery, formulae for power in Volt Amperes [VA] goes like this:

Power [volt-amperes]

For single-phase current:

Single-phase current

For three-phase current:

Three-phase current

The results of the power calculations are given in Table 1 below. The answers required are as follows.

1. The total input power kilovolt-amperes required for site is 120.8 kVA

2. The incoming supply cable capacity at 415 volt, three-phase, 50 Hz required is:


This is the input current to the site at the voltage of that cable. This is not the same as a total of the currents calculated in Table 1 as these larger numbers only appear at their reduced voltages in the relevant sub-circuits.

3. Assume that the crane, pump and tools are running for 25% of an 8-h working day, 5 days per week for 48 weeks, 20 of the tungsten lamps are for security lighting 16 h every night, and the remaining 40 tungsten lamps and the flood lamps are used for 3 h per day, 5 days per week for the winter period of 20 weeks.

The crane, pump and tools are working for:

Hours formula

The security lamps are working for:

Hours formula

The other lamps are working for:

Hours formula

Table 1 – Building site plant schedule

EquipmentPower (kW)NumberkWkVAVA
Tower Crane7517593.75415130.4
Sump Pump5156.2524026
Flood Lamps0.4124.84.824020
Hand Tools0.42081011090.9
Total //98.8120.8n.a.n.a.

Table 2 – Building site energy use

EquipmentPower (kW)NumberTime (h)Energy (kWh)
Tower Crane75148036000
Sump Pump514802400
Tungsten lamps, security Lamps0.120582411648
Flood Lamps0.4123001440
Hand Tools0.4204803840
Total Power Used //56528

The total energy used by the systems is found from:

kWh = number of appliances × kW per appliance × operation hours as shown in Table 2.

If electricity costs 8 p per unit (kWh) then the estimated cost for the 1-year contract will be:

Electricity cost

Reference: Building Services Engineering // Electrical Installations – David V. Chadderton (Purchase paperback from Amazon)

About Author //


Edvard Csanyi

Edvard - Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV high power busbar trunking (<6300A) in power substations, buildings and industry fascilities. Designing of LV/MV switchgears.Professional in AutoCAD programming and web-design.Present on


  1. magdy
    Aug 17, 2015

    we just start a construction of abig project at ajyad of mekkah saudi arabia we wanna make calculation of the total power needed for service using demand and load factor please tel me me how to compute the demand factor

  2. Munaf
    Aug 15, 2015

    Thank you for the information

  3. K.Ruju
    Mar 04, 2015

    Thanks for the Information

  4. jesus g
    Feb 21, 2015


  5. Loum Peter
    Feb 08, 2015

    I’m so sorry i had forgotten the principle of the power triangle so sorry really the 93.75kva for the crane is correct.

  6. Bimal
    Feb 08, 2015

    Nice good and very helpful

  7. Thomas
    Feb 07, 2015

    How about the power consumption for testing& commissioning?

  8. Loum Peter
    Feb 06, 2015

    Thanks for your information, but one think..for three phase , power W=1.732xV(line)xI(line)xPF no? if so that means VA =W/(1.732×0.8) and the KVA for the crane would be 54.13KVA….

    • jesus perez
      Feb 07, 2015

      No. Active power (W) is equal to VA x power factor

  9. sandar
    Feb 05, 2015

    Thanks u for ur information

  10. Anthony Uwadiae
    Feb 03, 2015

    Thanks for the information.

  11. Bonginkosi
    Feb 03, 2015

    Thank you Edvard Csanyi for an informative and summarized example.
    it assisted me in a project I am doing now.
    I keep reading your articles, these are very helpful articles
    the whole page is very useful, keep up the good work

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